Part 1: Fundamental Concepts: Thevenin and Norton Equivalence

In the previous section, a network of components, shown again in Figure 1, was reduced to a much simpler form. In that case, the goal was to find the voltage Vout. This was found with the understanding that Vout was not connected to anything else.

Schematic of the resistor network used to derive the Thevenin and Norton equivalent circuts.

Figure 1. A power source with an interconnected network of resistors.

What would you do if you were interested in knowing the effect of placing a resistive load on Vout? You could simply draw the additional resistor in the schematic, connect it to the Vout node, and simplify the circuit again. There is a more generalized approach that allows you to easily determine the effect of multiple load values on the circuit. This is where Thevenin’s and Norton’s Theorems come in. They state that a network of components can be replaced with an equivalent resistance, and either a voltage source in the case of Thevenin’s Theorem, or a current source in the case of Norton’s Theorem.

In both the Thevenin and Norton equivalent circuits, the open circuit voltage (the voltage with no load), and the voltage and current under load, will be equivalent to the original network with the same load. Note that, if the original group of components does not have a power source (a voltage or current source), these theorems do not apply. In that case you would simply combine the series and parallel values until the schematic was simplified.

Thevenin’s Theorem

Let’s take a look at Thevenin’s Theorem first. The theorem was developed by Hermann von Helmholtz in 1853 and independently by Léon Charles Thévenin in 1883. It states that any linear network, with current and voltage sources, can be replaced by a voltage source and a series resistor. A linear network is one where the relationship between the voltages and currents in the circuit remain the same regardless of their magnitudes. So far we have only concerned ourselves with Ohm's law, which is linear. It is linear because the relationship V = I*R does not changes as the values of the of V, I, and R change. Not all circuit elements behave this way. The following figure illustrates the concept.

Two schematics to illustraite the concept of a Thevenin equivalent circuit.

Figure 2. Thevenin equivalence concept.

When a load is placed across the connection points, A and B, in the left hand circuit and the right hand Thevenin equivalent, the same voltage will appear across it and the same current will flow through it no matter what value is chosen for the load. Although it might appear a bit daunting, finding values for Vth and Rth is straightforward.

There are any number of methods you can use to find Vth. In the previous section we found by simplifying the circuit that the output voltage is 1.558V. This turns out to be Vth. Vth is the voltage present across terminals A and B without a load. When there is no load, there is no current flow and therefore, no voltage drop across Rth. Simplifying the circuit to find the no load voltage yields the value Vth.

To find the value of Rth, called the resistance “looking into" the circuit, we need to replace the voltage source with a short circuit. Why do this? Recall that a voltage source does not change value regardless of the value of the load. This can only happen if a voltage source has no resistance. When we discussed voltage sources earlier, it was likened to a piece of wire across which a voltage appeared. Therefore, since we are only interested in Rth, we can replace the voltage source with a conductor. The following figure depicts the schematic with the voltage source replaced by a short.

Schematic illustrating the first step in developing the Thevenin equivalent.

Figure 3. Replacing the voltage source with a short..

The goal is to reduce the interconnected resistors to one equivalent value. This can be accomplished by simplifying the circuit. Looking at Figure 3, there are several immediate simplifications that can be made. R2 has a short (0Ω) in parallel with it so it has an equivalent resistance of 0Ω. R3 and R4 are connected in parallel so they can be combined into a single value. This is shown in the following figure.

Schematic illustrating the second step in developing the Thevenin equivalent.

Figure 4. Thevenin first simplification.

In Figure 4, R1, the 667Ω resistance, and R5 are in series with no branches leaving them so they can be combined into a single resistor. This is shown in the following figure.

Schematic illustrating the third step in developing the Thevenin equivalent.

Figure 5. Thevenin second simplification.

The next obvious simplification is to combine the 1767Ω resistor and R7. They are in parallel and are acting as a single resistance.

Schematic illustrating the fourth step in developing the Thevenin equivalent.

Figure 6. Thevenin third simplification.

We can now take the 639Ω resistance in series with R6 and take the resulting value in parallel with R8 to find the resistance across terminals A and B. This is shown in the figure below.

Figure 7. The Thevenin equivalent resistance.

We are almost done. We noted above that the open circuit voltage in the original schematic shown in Figure 1 was 1.558V, and that this value is the Thevenin equivalent voltage Vth. We just found the Thevenin equivalent resistance of 1,138Ω. Putting these two pieces of information together allows us to draw the Thevenin equivalent circuit. It is shown in the following figure along with the original schematic.

A schematic of the original circuit and its equivalent circuit with component values.

Figure 7. The Thevenin equivalent circuit.

The entire circuit shown on the left hand side of side of Figure 7, appears from the outside when viewed from, or looking into, its A and B terminals, to be a 1.558V voltage source with a 1,138Ω series resistance.

A couple of words of caution when dealing with Thevenin equivalent circuits. The equivalent circuit is valid and tells us how the original circuit will behave under load but it does not tell you much about the original circuit. For instance, the current flowing out of the 10V source in the original circuit with A and B shorted, may or may not be, the current flowing out of the 1.558V source in the equivalent circuit when A and B are shorted.

Norton’s Theorem

A Norton equivalent circuit is very similar to a Thevenin equivalent circuit except the voltage source is replaced with a current source and the series resistor is replaced with a parallel one. This is shown in the figure below using the same schematic used to demonstrate a Thevenin equivalent circuit.

Two schematics to illustraite the concept of a Norton equivalent circuit.

Figure 7. The Norton equivalent circuit.

In the case of the Norton equivalent circuit, the open circuit voltage is given by the current flow, In, through Rn and the short circuit current is given by In alone. If you spend some time thinking about the Thevenin and Norton equivalent circuits, you can begin to see how they represent the same thing. A voltage source has no internal resistance, electrically it looks like a piece of wire with a potential appearing across its ends. If you replace the voltage source with a short in a Thevenin equivalent circuit, looking into the terminal (A and B) you see the resistance Rth. A current source looks like an open circuit from which current flow. Removing the current source from the Norton equivalent yield a resistance looking into its terminal.

If you measure the open circuit voltage, there is no current flow in or out of the A and B terminals. No current flow means no voltage drop across the Thevenin resistance and no current division between an external load and the internal resistance in the Norton equivalent.

When you develop the Norton equivalent circuit, you can begin from the original circuit and derive the equivalent circuit or you can start with the Thevenin equivalent and apply a few simple equations. This is the approach taken here.

From the previous discussion, we understand that the Thevenin series resistance is equivalent to the Norton parallel resistance.

The value of the Norton equivalent current source can be found by considering the Thevenin open circuit voltage and the internal resistance. Ohm's law applies here and the value of the current source is found by using the following equation.

Equation 1. Norton current value.

Equation 1 finds the current that must flow through Rn so that the open circuit voltage is Vth, the no load terminal voltage. Using Equation 1 and the fact that the Norton and Thevenin equivalent resistances are the same, it is possible to take a Thevenin equivalent circuit and produce a Norton equivalent circuit with the same characteristics.

The following figure shows the Thevenin equivalent circuit from Figure 1 and a Norton equivalent circuit derived from the same original schematic.

Schematics of the Thevenin and Norton equivalent circuits for the same network.

Figure 7. The Norton and Thevenin equivalent circuit.

Let’s see how these equivalent circuit hold up by applying some loads to them. The first thing to try is how the open circuit voltages compare. For the Thevenin equivalent this is trivial. With no load applied, there is no current flow, so the terminal voltage is 1.558V. Using Ohm's law gives us the no load voltage for the Norton equivalent. With no current flowing from its terminals, all of the current flows through the internal resistance which gives a voltage of V = 0.001369 * 1138 or 1.558V.

Next, let’s see how the short circuit currents compare. With a short across the terminals of the Thevenin circuit, it becomes a series circuit with the current limited by the internal resistance. Using Ohm's law we find that I = 1.558/1138 or I = 0.001369. With a short across the terminals of the Norton circuit, we end up with series circuit agin. With a short, or 0Ω connection, across A an B, all of the current end up flowing through it. A 0Ω resistor in parallel with any other resistance yields a net resistance of 0Ω. As a result, the short circuit current in the Norton equivalent circuit is the current source value of 0.001369.

Finally, let’s try a randomly chosen load resistance of 500Ω. In the Thevenin equivalent circuit we end up with a voltage source in series with two resistors. The total current is found by I = 1.558/(1138 + 500) which yields a current of 0.0009512A. This current flowing through the 500Ω yields the voltage across the load of V = 0.0009512 * 500 or 0.476V. In the case of the Norton equivalent, we end up with a current source in parallel with two resistor. The value of the resistors in parallel is 347.37Ω. The voltage across the load is therefore due to the fixed current from the current source flowing though the parallel resistance of 347.37Ω which act a single component. Ohm's law yields V = 0.001369 * 347.47 or V = 0.476V.

You could continue to try applying different loads to the Thevenin and Norton equivalents if you want, but the result will always be the same. The results will match one another and yield the same results if you were to apply the same loads to the original circuit.

There are very few occasions where you will use these theorems when working with circuits. That said, the concepts involved are worth remembering. They form part of the collection of ideas that allow you to understand how circuits work.

One of the few practical applications has to do with interfacing to the circuit. You may be connecting to a black box or adding circuitry to an existing schematic. Having the simplified equivalent circuit makes that task much simpler. You don’t need to analyze the existing circuit, you only need a simple description of it to complete what you are doing. This is also useful when working as part of a team where each member of the team is tasked with developing a different part of the system.